Question

The pressure at the bottom of a cylindrical container with a
cross-sectional area of 58.5 cm^{2} and holding a fluid of
density 500 kg/m^3 is 115 kPa.

(a) Determine the depth of the fluid.

(b) Determine the pressure at the bottom of the container if an
additional 1.80 10^{-3} m^{3} of this fluid is
added to the container. (Give your answer to at least 3 significant
figures.)

Answer #1

**given
cross sectional area, A = 58.5 cm^2
rho = 500 kg/m^3
P = 115 kPa**

**= 1.15*10^5 pa**

**a) let h is the depth of the fluid**

**we know, P = P_atm + rho*g*h**

**rho*g*h = P - P_atm**

**h = (P - P_atm)/(rho*g)**

**= (1.15*10^5 - 1.0134*10^5)/(500*9.8)**

**= 2.79 m
<<<<<<<<<<<-------------------Answer**

**b) New depth of the cyllinder, H = h + Volume
added/A**

**= 2.79 + 1.80*10^-3/(58.5*10^-4)**

**= 3.098 m**

**P = P_atm + rho*g*h**

**= 1.0134*10^5 + 500*9.8*3.098**

**= 1.165*10^5 pa (or) 116.5 kPa
<<<<<<<<<---------------Answer**

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